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I'm reading the Art of electronics and trying to calculate current and voltages of the following scheme:

schematic

simulate this circuit – Schematic created using CircuitLab

I wrote the current as the letters with indexes (I1-I9).

So I got the following equations:

I1 = I5 + I6
I2 = I3 + I4
I9 = I6 + I4
I8 = I7 + I5

Also we know that:

I1 = 5/1000 = 0.005A
I2 = 5/1000 = 0.005A

Then I tried calculating the first transistor's current (on the left side):

I7 = V/R = (5 - 0.6)/25K = 0.000176A

Then I used the general rule that:

Ie = b * Ib, where b=beta coefficient, Ie - emitter current, Ib - base current

So we got something like:

I8 = b * I7

Lets say b = 100, then

I8 = 0.000176 * 100 = 0.0176A

From I8 = I5 + I7 we get that:

0.0176 = 0.000176 + I5

giving us:

I5 = 0.017424A

I1 = I5 + I6:

0.005 = 0.017424 + I6

I6 = -0.012424A

And here I don't understand why its minus. Whenever transistors and diodes comes into the circuits I can't estimate the voltages and the currents. I think that I'm missing something there.

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    \$\begingroup\$ If you look at the circuit the base-emitter junction of Q2 is clamping the collector-emitter of Q1 at 0.6V so I1 = 4.4/1(k) = 4.4mA \$\endgroup\$ – JIm Dearden Jul 25 '16 at 10:53
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    \$\begingroup\$ Also we know that: I1 = 5/1000 = 0.005A I2 = 5/1000 = 0.005A - How do you know that? That is only true if the other end of those resistors is at 0V. \$\endgroup\$ – Andrew Jul 25 '16 at 11:18
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You are forgetting that you can overdrive a base so that it goes into saturation. This is desirable to get the transistor fully turned on.

Your calculation for I7 is correct:

$$ I_7 = \frac {V}{R_3} = \frac {5 - 0.6}{25k} = 0.176~mA $$

You then calculate the maximum collector current using \$ I_8 = \beta I_7 \$ Lets say \$ \beta \$ = 100, then \$ I_8 = 0.176 * 100 = 17.6~mA \$.

Now look at R1. The maximum current through it is given by

$$ I_1 = \frac {V}{R_1} = \frac {5 - 0.6}{1k} = 4.4~mA $$

Clearly the maximum possible value for \$ I_8 \$ is this 4.4 mA + 0.176 mA from \$ I_7 \$.

The calculations indicate that we could reduce the base current into Q1. In practice we want to ensure that the drive is adequate to cover variations in \$ \beta \$ due to production spread or even transistor substitution. Drive it hard and make sure it's fully 'on' is the normal approach.

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You can only use beta as you have if you can be sure that there is enough "drive potential" at the collector to produce the amplified base current. Clearly the maximum current through R1 is 5 mA and this means that the base current x 100 (beta) has to be somewhat less than 5 mA.

Beta is decimated when the transistor starts to hit saturation.

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